It requires 10.15 kilojoules of energy.
What is vaporization?The term "vaporisation" (or "evaporation") often refers to the transformation of a liquid's condition into a vapour phase below its boiling point. The phrase, however, can also refer to the process of removing a solvent, independent of the temperature used.
What is energy?When a body moves to exert force, it is said to be exerting work. Energy is the capacity to accomplish work. Energy is something we always need, and it can take many different forms.
If the gold is present in the liquid state, you only have to determine the latent heat of vaporization, or lvap. The empirical data for gold is 330 kJ/mol.
Q = mlvap
Q = (2 kg)(1 kmol/197 kg)(1,000 mol/1 kmol)
Q = 10.15 kJ
It needs an energy of 10.15 kilojoules
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What is eutectic temperature
The eutectic point is the lowest temperature at which the liquid phase is constant at a particular pressure.
What does the word "eutectic" mean?A melting composition known as a eutectic consists of at least two components that melt and freeze at the same rates. The components combine during the crystallisation phase, operating as a single component as a result.
What are eutectic pressure and temperature?The eutectic is the system's lowest melting point under its own pressure; it has a matching temperature called the eutectic temperature and produces the eutectic liquid as a result. In terms of composition, eutectic liquids are located between the system's solid phases.
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pls help!!!
a compound is found to be 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen. it has a molecular molar mass of 140.22 g/mol. what is the molecular formula.
show work pls!!
The molecular formula of the compound, given that it contains 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen is C₆H₁₂N₄
How do i determine the molecular formula?To obtain the molecular formula, we must first determine the empirical formula. Details on how to obtain the empirical formula is given beloww:
Carbon (C) = 51.39%Hydrogen (H) = 8.64%Nitrogen (N) = 39.97%Empirical formula =?Divide by their molar mass
C = 51.39 / 12 = 4.283
H = 8.64 / 1 = 8.64
N = 39.97 / 14 = 2.855
Divide by the smallest
C = 4.283 / 2.855 = 1.5
H = 8.64 / 2.855 = 3
N = 2.855 / 2.855 = 1
Multiply through by 2 to express in whole number
C = 1.5 × 2 = 3
H = 3 × 2 = 6
N = 1 × 2 = 2
Thus, we can conclude that the empirical formula is C₃H₆N₂
Finally, we shall determine the molecular formula. Details below
Empirical formula = C₃H₆N₂Molar mass of compound = 140.22 g/molMolecular formula =?Molecular formula = empirical × n = mass number
[C₃H₆N₂]n = 140.22
[(12×3) + (1×6) + (14×2)]n = 140.22
70n = 140.22
Divide both sides by 70
n = 140.22 / 70
n = 2
Molecular formula = [C₃H₆N₂]n
Molecular formula = [C₃H₆N₂]₂
Molecular formula = C₆H₁₂N₄
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Consider the reaction described by the chemical equation shown.
C2H4(g)+H2O(l)⟶C2H5OH(l)Δ∘rxn=−44.2 kJ
Use the data from the table of thermodynamic properties to calculate the value of Δ∘rxn
at 25.0 ∘C.
ΔS∘rxn= ? J⋅K−1
Calculate Δ∘rxn.
ΔG∘rxn= ? kJ
In which direction is the reaction, as written, spontaneous at 25 ∘C
and standard pressure?
reverse
both
neither
forward
Answer:
To calculate Δ∘rxn, we can use the following formula:
ΔG∘rxn = ΔH∘rxn - TΔS∘rxn
where ΔH∘rxn is the enthalpy change of the reaction, T is the temperature in Kelvin, and ΔS∘rxn is the entropy change of the reaction.
We know that ΔH∘rxn = -44.2 kJ and we want to find ΔS∘rxn at 25.0 ∘C (298 K). We can use the following formula to calculate ΔS∘rxn:
ΔG∘rxn = -RTlnK
where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and K is the equilibrium constant.
We can find K using the following formula:
ΔG∘rxn = -RTlnK K = e^(-ΔG∘rxn/RT)
We know that ΔG∘rxn = -44.2 kJ/mol and R = 8.314 J/mol K, so we can calculate K:
K = e^(-(-44.2 kJ/mol)/(8.314 J/mol K * 298 K)) K = 1.9 x 10^7
Now we can use K to calculate ΔS∘rxn:
ΔG∘rxn = -RTlnK ΔS∘rxn = -(ΔH∘rxn - ΔG∘rxn)/T ΔS∘rxn = -((-44.2 kJ/mol) - (-8.314 J/mol K * 298 K * ln(1.9 x 10^7)))/(298 K) ΔS∘rxn = -0.143 kJ/K
Therefore, ΔS∘rxn is -0.143 kJ/K.
To determine whether the reaction is spontaneous at 25 ∘C and standard pressure, we can use Gibbs free energy (ΔG). If ΔG < 0, then the reaction is spontaneous in the forward direction; if ΔG > 0, then it is spontaneous in the reverse direction; if ΔG = 0, then it is at equilibrium.
We know that ΔG∘rxn = -44.2 kJ/mol and T = 25 ∘C (298 K). We can use the following formula to calculate ΔG:
ΔG = ΔG∘ + RTlnQ
where Q is the reaction quotient.
At equilibrium, Q = K (the equilibrium constant). Since we calculated K earlier to be 1.9 x 10^7, we can use this value for Q.
ΔG = ΔG∘ + RTlnQ ΔG = (-44.2 kJ/mol) + (8.314 J/mol K * 298 K * ln(1.9 x 10^7)) ΔG = -43.6 kJ/mol
Since ΔG < 0, the reaction is spontaneous in the forward direction at 25 ∘C and standard pressure.
2. A student prepared a 0.500 M solution of an unknown acid, and measured the pH as 3.56 at 25°C. (a) What is the acid dissociation constant of this unknown acid? (b) What percentage of acid is ionised in this solution
To solve this problem, we can use the following equation that relates the pH of a solution to the acid dissociation constant (Ka) and the concentration of the acid:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in the solution.
(a) To find the Ka of the unknown acid, we need to first find the concentration of hydrogen ions in the solution. We can do this by taking the inverse of the pH and converting it to a concentration:
[H+] = 10^(-pH) = 10^(-3.56) = 2.17 × 10^(-4) M
What is the acid dissociation constant of this unknown acid?The acid dissociation constant (Ka) can then be calculated using the equation:
Ka = [H+][A-]/[HA]
where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the undissociated acid. Since we don't know the values of these concentrations, we need to use the fact that the solution is 0.500 M to make an assumption about the degree of dissociation (α) of the acid:
α = [A-]/[HA]
Since the solution is not extremely dilute, we can assume that the degree of dissociation is small and that the concentration of the undissociated acid is approximately equal to the initial concentration of the acid. Therefore, we can write:
[A-] ≈ 0.500α
[HA] ≈ 0.500 - 0.500α
Substituting these expressions into the equation for Ka, we get:
Ka = [H+][A-]/[HA] ≈ ([H+][A-])/0.500α
≈ ([H+]/Ka)(0.500α)/(1-α)
Solving for Ka, we get:
Ka ≈ H+/0.500α
Substituting the values we have calculated, we get:
Ka ≈ (2.17 × 10^(-4))(1-α)/(0.500α) = 4.37 × 10^(-5)
Therefore, the acid dissociation constant of the unknown acid is approximately 4.37 × 10^(-5).
(b) To find the percentage of acid that is ionized in the solution, we can use the equation:
α = [A-]/[HA] = 10^(-pKa + pH)/(1 + 10^(-pKa + pH))
where pKa is the negative logarithm of the acid dissociation constant. Substituting the values we have calculated, we get:
α = 10^(-(-4.36) + 3.56)/(1 + 10^(-(-4.36) + 3.56)) ≈ 0.008
Therefore, the percentage of acid that is ionized in the solution is approximately 0.8%.
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To solve this problem, we can use the following equation that relates the pH of a solution to the acid dissociation constant (Ka) and the concentration of the acid:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in the solution.
(a) To find the Ka of the unknown acid, we need to first find the concentration of hydrogen ions in the solution. We can do this by taking the inverse of the pH and converting it to a concentration:
[H+] = 10^(-pH) = 10^(-3.56) = 2.17 × 10^(-4) M
What is the acid dissociation constant of this unknown acid?The acid dissociation constant (Ka) can then be calculated using the equation:
Ka = [H+][A-]/[HA]
where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the undissociated acid. Since we don't know the values of these concentrations, we need to use the fact that the solution is 0.500 M to make an assumption about the degree of dissociation (α) of the acid:
α = [A-]/[HA]
Since the solution is not extremely dilute, we can assume that the degree of dissociation is small and that the concentration of the undissociated acid is approximately equal to the initial concentration of the acid. Therefore, we can write:
[A-] ≈ 0.500α
[HA] ≈ 0.500 - 0.500α
Substituting these expressions into the equation for Ka, we get:
Ka = [H+][A-]/[HA] ≈ ([H+][A-])/0.500α
≈ ([H+]/Ka)(0.500α)/(1-α)
Solving for Ka, we get:
Ka ≈ H+/0.500α
Substituting the values we have calculated, we get:
Ka ≈ (2.17 × 10^(-4))(1-α)/(0.500α) = 4.37 × 10^(-5)
Therefore, the acid dissociation constant of the unknown acid is approximately 4.37 × 10^(-5).
(b) To find the percentage of acid that is ionized in the solution, we can use the equation:
α = [A-]/[HA] = 10^(-pKa + pH)/(1 + 10^(-pKa + pH))
where pKa is the negative logarithm of the acid dissociation constant. Substituting the values we have calculated, we get:
α = 10^(-(-4.36) + 3.56)/(1 + 10^(-(-4.36) + 3.56)) ≈ 0.008
Therefore, the percentage of acid that is ionized in the solution is approximately 0.8%.
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When a hydrogen atom is added to a polyatomic ion, the amount of negative charge . Following this pattern, we can see that hydrogen carbonate has a charge of and hydrogen sulfate has a charge of .
If we add one or two hydrogen ions to a polyatomic ion that has a 3-charge, as the phosphate ion (PO₄3-), it will still be a polyatomic ion. (Three H+ would entirely cancel out the 3-charge, turning it into a neutral molecule and removing it from the category of polyatomic ions.
Why does carbonate have a negative 2 charge?As a result, the carbonate ion has 2 more electrons than protons due to its negative charge. The doubly bonded oxygen in the carbonate ion is neutral, whereas each single bonded oxygen has a negative charge. This is the cause of the total charge of "-2," then.
An essential component of the atmosphere of stars like the Sun is the hydrogen anion.
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The temperature of a 2.0-liter sample of helium gas at STP is increased to 27C, and the pressure is decreased to 80 kPa. What is the new volume of the helium sample? Round your answer to the nearest tenth of a liter?
The new volume of the helium sample would be 2.4 L.
Volume of a gasAccording to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvins.
At STP (standard temperature and pressure), which is defined as 0°C (273.15 K) and 101.325 kPa, the volume of 2.0 liters of helium gas contains one mole of helium atoms.
To find the new volume of the helium sample when the temperature is increased to 27°C (300.15 K) and the pressure is decreased to 80 kPa, we can use the following equation:
(P1V1)/T1 = (P2V2)/T2
where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.
Plugging in the values, we get:
(101.325 kPa)(2.0 L)/(273.15 K) = (80 kPa)(V2)/(300.15 K)
Solving for V2, we get:
V2 = (101.325 kPa)(2.0 L)/(273.15 K) * (300.15 K)/(80 kPa) = 2.36 L
Therefore, the new volume of the helium sample is approximately 2.4 L (rounded to the nearest tenth).
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Round to 2 significant
figures.
5,249
5,250. The number was rounded up from 5,249 because the last digit, 9, is greater than or equal to 5.
What is rounded up?Rounding up is a mathematical operation that involves increasing a number to its nearest whole number. It is commonly used when dealing with money, measurements, or statistics. When rounding up, the number is increased to the next highest whole number. For example, if a number is 6.7, it would be rounded up to 7. Rounding up is often used when dealing with exact measurements or estimates to simplify the calculations. It can also be used to make the results of a calculation easier to understand. In the case of money, rounding up can be used to round a number to the nearest dollar. This prevents dealing with fractional amounts of money. Rounding up can also be utilized in statistical analysis, such as in the calculation of mean or median. This simplifies the data and prevents dealing with fractions or decimals.
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If all the coefficients in the already balanced equation are multiplied by 2, will the equation still remain balanced and will the multiplication affect the equilibrium constant? If you answer yes to any part of the question, please explain in detail.
Yes, If all the coefficients in a balanced chemical equation are multiplied by 2, the equation will still remain balanced because the ratio of the reactants and products remain the same.
What are the coefficients?
For example, the balanced equation: [tex]2H_{2}[/tex] + [tex]O_{2}[/tex] -> [tex]2H_{2}O[/tex]
The coefficients of a chemical equation are the numbers written in front of the chemical formulas of reactants and products, indicating the relative amounts of each substance involved in the reaction. The coefficients are used to balance the chemical equation, ensuring that the law of conservation of mass is obeyed. The coefficients represent the smallest whole-number ratios of the substances in the reaction, and they provide important information about the stoichiometry of the reaction.
When all the coefficients are multiplied by 2, the equation becomes: [tex]4H_{2}[/tex] + [tex]2O_{2}[/tex] -> [tex]4H_{2}O[/tex]
The equation is still balanced because the ratio of hydrogen to oxygen to water molecules remains the same (4:2:4).
Multiplying the coefficients by a constant will not affect the equilibrium constant (Kc) as long as the reaction conditions remain constant. This is because the equilibrium constant is a ratio of the concentrations of products and reactants at equilibrium, and the ratio remains the same even if the coefficients of the balanced equation are multiplied by a constant. However, if the temperature, pressure or concentration of any reactants or products are changed, then the value of the equilibrium constant will change.
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6. What is the pH of a 0.25 M solution of NH4Cl? [Kb(NH3) = 1.8 10–5
The Ammonium Chloride solution at 0.25 M has a pH of 2.67.
Why is the pH of Ammonium Chloride below 7?As a result, the weak basic (Chlorine) in the solution is overpowered by the conjugate acid (Ammonium cation), making the solution mildly acidic. According to the equation pH =log[Hydrogen ion], an acidic solution has a pH lower than 7. Aqueous ammonium chloride solution has a pH that is less than 7.
Ammonium cation + Water ⇌ Nitrogen trihydride + Hydronium ion
Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]
[Nitrogen trihydride] = [Hydronium ion] = x
[Ammonium cation] = 0.25 - x
Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]
1.8 × 10–5 = x² / (0.25 - x)
1.8 × 10–5 = x² / 0.25
x² = 4.5 × 10–6
x = 2.12 × 10–3
pH = -log[Hydronium ion] = -log(2.12 × 10–3) = 2.67
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Question 5(Multiple Choice Worth 3 points)
(07.02 LC)
The substances below are listed by increasing specific heat capacity value. Starting at 30.0 °C, they each absorb 100 kJ of thermal energy. Which one do you expect to increase in temperature the least?
a) Cadmium, 0.230 J/(g °C)
b) Sodium, 1.21 J/(g °C)
c) Water, 4.184 J/(g °C)
d) Hydrogen, 14.267 J/(g °C)
Component form of the vector v is as follows: 4 3 1.5 1 Using the standard basis vectors I and j), express the vector w as follows: 3 two 1 4 pp . 1 3 w 3.5 C. V plus w= d. Determine the vector v's magnitude
What does "vector" mean?
Latin word for "carrier" is "vector." Point A is transported to point B by vectors. The orientation of the vectors AB is the direction in which point A is moved in relation to point B, and the amplitude of the vector is the width of the line connecting the two locations A and B. The terms Euclidean vectors and spatial vectors are also used to refer to vectors.
A vector space is what?
A vector space, also known as a linear space, is a collection of things called vectors that can be added to and multiplied ("scaled") by figures called scalars in the fields of mathematics, physics, and engineering.
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Sulfurous acid ( H2SO3) is a diprotic acid with a1=1.39×10−2 and a2=6.73×10−8.
Determine the pH of a 0.189M sulfurous acid ( H2SO3 ) solution.
A sulfuric acid solution with a concentration of 0.189M has a pH of roughly 0.778.
Sulfuric acid: Is it a diprotic acid?A diprotic acid is sulfuric acid, Sulfurous acid. The findings of an acid-base titration can be used to calculate the values of Ka1 and Ka2. To completely neutralise diprotic acids, two hydroxide ions are needed for each molecule. Sulfuric acid, a stronger acid than Hydrogen sulfide, with a pH of 1.5 in a solution of 0.100 mol dm³.
The chemical equation for sulfuric acid dissociation is as follows:
Sulfuric acid + Water ⇌ Hydronium ion + hydrosulfite anion
Write the expression for the acid dissociation constant (Ka1) for the first dissociation:
Ka1 = [Hydronium ion[hydrosulfite anion]/[Sulfuric acid]
Substitute the given value of Ka1 (1.39×10−2) and the initial concentration of Sulfuric acid (0.189 M) into the expression for Ka1 and solve for [hydronium ion]:
1.39×10−2 = [hydronium ion][Hydrogen sulfite]/0.189
[hydronium ion] = 0.167 M
pH = -log[hydronium ion]
pH = -log(0.167)
pH = 0.778
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50 points +brainlist (there's going to be 3 more added on my profile with the same points(
which type of process is this?
chemical
physical
nuclear
nuclear type of process is this
Is the reaction physical or chemical?The content of a physical reaction differs from that of a chemical reaction. A chemical reaction changes the makeup of the substances in question; a physical change changes the look, smell, or plain presentation of a sample of matter without changing its content.
Nuclear reactions are not the same as chemical reactions. Atoms become more stable in chemical processes by engaging in electron transfers or by sharing electrons with other atoms.
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whats the answer and why?
I would say C
Since the nitro group (NO2) contains a positively charged nitrogen atom, it tends to attract electron from the aromatic ring and, therefore, the other group/atom. In the first case, I think piridine (II) makes a stronger bond with water since the nitrogen in the aromatic ring needs its electrons in order to be have a slight negative charge that can interact with the slightly positive charged hydrogen atom in water. If the nitro group is present, it will attract to some extent the electrons of the nitrogen atom in the ring, thus making the H-bond less stronger.
In the second case the hydrogen, which is slightly positive, of the OH group interacts with the oxygen, which is slightly negative, of water. If the nitro group is present, it will attract the electrons of oxygen of the hydroxyl group, therefore making the bond between the oxygen and the hydrogen more polar (which basically means that the bonding electron of hydrogen is even more attracted by the oxygen atom) making the hydrogen atom more positive, which means that the H-bond will be stronger
Which sub atomic particles are similar in size
Answer:
Neutrons and Protons
Explanation:
Different elements can have subatomic particles of varying sizes. The size of an atom is defined by the size of its electron cloud, which is composed of electrons, and the size of its nucleus, which is composed of protons and neutrons. The atomic number and subsequently the identity of an element are determined by the number of protons in the nucleus. The quantity of protons and neutrons in the nucleus determines its size. The quantity of electrons in the electron cloud and the energy levels they are located at define its size. The size of atoms can differ depending on the element due to differences in the amount of protons, neutrons, and electrons.
Select all the elementary substances.
silver bromide (AgBr)
silicon dioxide (SiO₂)
hydrogen sulfide (H₂S)
xenon (Xe)
Answer:
silicon dioxide,xenon
Explanation:
What happens when a solid is dissolved into a liquid?
.
Which of the following represents beta decay
OA. Tc-TC+y
O B.
B. 14Gd→ 144Sm+ He
O C. 160Eu+e→ 169 Sm
62
O D.
D.
63
164Gd→ ¹6 Tb + e
160
65
The correct answer that represents beta decay is
D. 164Gd → 164Tb + e, What happens in beta decayIn beta decay, a neutron in the nucleus is converted into a proton, and an electron (or beta particle) and an antineutrino are emitted from the nucleus.
In this case, a neutron in the 164Gd nucleus is converted into a proton, and an electron is emitted from the nucleus, resulting in the production of 164Tb.
Option A is not a valid representation of any known type of radioactive decay.
Option B represents alpha decay, in which an alpha particle is emitted from the nucleus.
Option C represents electron capture, in which an electron is captured by the nucleus.
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Draw both enantiomers of the following compound
Enantiomers rotate the plane of polarized light in opposite directions, and this property is used to distinguish between them in a process called optical rotation.
What are the enantiomers of a compound?Enantiomers are pairs of molecules that are non-superimposable mirror images of each other.
They are isomers, meaning they have the same molecular formula and connectivity but differ in their three-dimensional arrangement of atoms in space.
Enantiomers exhibit identical physical and chemical properties, except for their interaction with plane-polarized light (a type of light that oscillates in a single plane).
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Given the following data for water:
Heat of fusion = 334 J/g
Heat of vaporization = 2,256 J/g
Specific heat of solid = 2.09 J/g °C)
Specific heat of liquid = 4.184 J/g °C)
Specific heat of gas = 1.84 J/g °C)
Calculate how much energy is needed to change 100.0 grams of liquid water at 15.0 °C to vapor at 125.0 °C. (3 points)
Oa
O
b
44,000 J
89,400 J
104,000 J
266,000 J
The Ka value for ethanoic acid, CH3COOH is 1.79 x 10-5. What is the pH of an equimolar solution of ethanoic acid and Na+CH3COO-?
The pH of the solution can be calculated using the following steps:
Write the chemical equation for the dissociation of ethanoic acid:
CH3COOH + H2O ⇌ CH3COO- + H3O+
Write the equilibrium expression for the dissociation of ethanoic acid:
Ka = [CH3COO-][H3O+] / [CH3COOH]
Since the solution is equimolar in CH3COOH and CH3COO-, we can assume that the initial concentrations of CH3COOH and CH3COO- are equal. Let's use the variable x to represent the concentration of CH3COO- and CH3COOH in mol/L.
[CH3COOH] = x mol/L [CH3COO-] = x mol/L
Since CH3COOH is a weak acid, we can assume that only a small fraction of it dissociates in water. Let's use the variable y to represent the concentration of H3O+ ions in mol/L that are produced from the dissociation of CH3COOH. From the dissociation of ethanoic acid, we know that [CH3COO-] = [H3O+].
[CH3COO-] = y mol/L [H3O+] = y mol/L
Use the equilibrium expression to solve for the concentration of H3O+ ions:
Ka = [CH3COO-][H3O+] / [CH3COOH] 1.79 x 10^-5 = y^2 / x
Solving for y in terms of x, we get:
y = sqrt(Ka * x)
Calculate the pH of the solution using the equation:
pH = -log[H3O+]
pH = -log(y)
Substituting in the value of y from Step 5, we get:
pH = -log(sqrt(Ka * x))
Simplifying, we get:
pH = -0.5 * log(Ka * x)
Substituting in the value of Ka, we get:
pH = -0.5 * log(1.79 x 10^-5 * x)
Now we can calculate the pH for the solution by substituting the value of x as it is equimolar.
pH = -0.5 * log(1.79 x 10^-5 * x)
pH = -0.5 * log(1.79 x 10^-5 * 1)
pH = -0.5 * log(1.79 x 10^-5)
pH = 4.74
Therefore, the pH of an equimolar solution of ethanoic acid and Na+CH3COO- is 4.74.
At 25 ∘C
, the equilibrium partial pressures for the reaction
A(g)+2B(g)↽−−⇀C(g)+D(g)
were found to be A=5.63
atm, B=5.00
atm, C=5.47
atm, and D=5.63
atm.
What is the standard change in Gibbs free energy of this reaction at 25 ∘C
?
The standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.
What is standard change?
To find the standard change in Gibbs free energy of the reaction, we need to use the following equation:
ΔG° = -RT ln(K)
where ΔG° is the standard change in Gibbs free energy, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25 °C = 298 K), and K is the equilibrium constant.
To find K, we need to use the equilibrium partial pressures:
K = (PC × PD) / (PA × PB²)
where PA, PB, PC, and PD are the equilibrium partial pressures of A, B, C, and D, respectively.
Substituting the values, we get:
K = (5.47 atm × 5.63 atm) / (5.63 atm × (5.00 atm)²)
K = 0.6176
Now we can calculate the standard change in Gibbs free energy:
ΔG° = -RT ln(K)
ΔG° = -(8.314 J/mol·K) × (298 K) × ln(0.6176)
ΔG° = -1,690 J/mol or -1.69 kJ/mol
Therefore, the standard change in Gibbs free energy of the reaction at 25 ∘C is -1.69 kJ/mol.
What is free energy?
Free energy, also known as Gibbs free energy, is a thermodynamic quantity that represents the amount of energy in a system that is available to do work at a constant temperature and pressure. It is denoted by the symbol G and is expressed in units of joules (J) or calories (cal).
In simple terms, free energy is the energy that can be used to do work. It is defined by the equation:
ΔG = ΔH - TΔS
where ΔH is the change in enthalpy (heat content) of the system, ΔS is the change in entropy (disorder) of the system, and T is the absolute temperature in Kelvin.
If ΔG is negative, the reaction is spontaneous and can proceed without the input of external energy. If ΔG is positive, the reaction is non-spontaneous and requires energy input to proceed. If ΔG is zero, the system is at equilibrium.
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Calcium nitrate reacts with ammonium fluoride to make calcium fluoride and ammonium nitrate. When (4.479x10^1) mL of (4.61x10^-1) M calcium nitrate was added to (7.332x10^1) mL of (1.5835x10^0) M ammonium fluoride, 0.731 grams of calcium fluoride were isolated. How many moles of ammonium fluoride were initially added in this experiment (not necessarily reacted)?
The moles of ammonium fluoride initially added in this experiment was 0.0216 moles.
What is mole?Mole is a unit of measurement that is used in chemistry to measure the amount of a substance. It is a very important unit of measurement because it allows chemists to accurately measure the amount of a substance that is being used in a reaction. The mole is defined as the amount of a substance that contains the same number of particles as there are atoms in 12 grams of carbon-12..
First, we need to calculate the moles of calcium nitrate in the solution. We can do this by using the molarity and volume of the solution:
(4.61x10⁻¹ M)*(4.479x10¹ mL) = 0.0216 moles of calcium nitrate
(0.731 g)*(1 mol/55.847 g) = 0.0131 moles of calcium fluoride
(0.0216 moles)*(1 mol/1 mol)
= 0.0216 moles of ammonium fluoride
Therefore, the moles of ammonium fluoride initially added in this experiment was 0.0216 moles.
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2. When dinitrogen pentoxide is heated, it decomposes to
nitrogen dioxide and oxygen. How many moles of nitrogen
dioxide can be formed from the decomposition of 1.25 g of
dinitrogen pentoxide?
0.02314 moles of NO₂ can be formed from the decomposition of 1.25 g of dinitrogen pentoxide.
The balanced equation for the decomposition of dinitrogen pentoxide is:
2 N₂O₅ → 4 NO₂ + O₂
The molar mass of N₂O₅ is 108.01 g/mol.
To determine the number of moles of N₂O₅ present in 1.25 g, we use the following calculation:
moles N₂O₅ = mass / molar mass
moles N₂O₅ = 1.25 g / 108.01 g/mol
moles N₂O₅ = 0.01157 mol
From the balanced equation, we can see that 2 moles of N₂O₅ decompose to form 4 moles of NO2. Therefore, the number of moles of NO2 produced can be calculated as:
moles NO₂ = (0.01157 mol N2O5) × (4 mol NO2 / 2 mol N2O5)
moles NO₂ = 0.02314 mol
Therefore, 0.02314 moles of NO₂ can be formed from the decomposition of 1.25 g of dinitrogen pentoxide.
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CHALLENGE The circles below represent of the large circle, and multiply it by 30. That Earth and the moon. Measure the diameter would be the correct distance from Earth to the moon at this scale. Draw the two circles in the space provided. Use the correct distance you found.● = Earth ●=moon
To draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.
What is the explanation for the above response?If we assume that the larger circle represents the Earth, then the diameter of the Earth would be 30 times the diameter of the smaller circle representing the moon. Let's say that the diameter of the smaller circle is x. Then the diameter of the larger circle (Earth) would be 30 times x or 30x.
To find the correct distance from Earth to the moon at this scale, we need to know the actual distance from Earth to the moon, which is approximately 238,855 miles or 384,400 kilometers. If we divide this distance by the scale factor of 30, we get:
238,855 miles / 30 = 7,961.8 miles
Therefore, the diameter of the smaller circle (moon) would be approximately 7,961.8 miles / π = 2,532.5 miles (rounded to one decimal place). And the diameter of the larger circle (Earth) would be 30 times that or 75,974.4 miles
So, to draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.
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Which statement about the reaction between calcium oxide and water is correct?
a) 65.2 kJ of heat are released for every mole of CaO that reacts.
b) 130 kJ of heat are released for every mole of H2O that reacts.
c) 130 kJ of heat are absorbed for every mole of CaO that reacts.
d) 65.2 kJ of heat are absorbed for every mole of H2O that reacts.
The heat that is released is 1600 J
65.2 kJ of heat are released for every mole of CaO that reacts.
What is the heat released?For every mole of CaO that combines with water, 65.2 kJ of heat are generated, according to thermodynamic statistics. As a result, a considerable amount of heat is emitted throughout the reaction, making it highly exothermic.
We know that;
H = mcdT
H = heat absorbed or evolved
m = mass of the substance
c = Heat capacity of the substance
dT = temperature change.
H = 60 * 1 * (43 - 70)
H = -1600 J
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What does X represent for this transmutation? 9 4Be + 4₂He X+ ¹on ?
The result of the transformation, denoted by the symbol X, is 12 6C.
What does the radioactive decay symbol X stand for?The chemical symbol for the unstable nucleus, X, is represented by the nuclear equation, where the letter a stands for the particle's mass number and the letter b for the number of protons.
What is atom transmutation?the process of changing one chemical element into another. Since a transmutation involves a change to the atomic nuclei's structure, it can either be produced via a nuclear reaction (q.v. ), like neutron capture, or it can happen naturally due to radioactive decay, like alpha and beta decay (qq. v.).
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8. Balance the following equation:
NH3(g) + F2(g) → N₂F4(g) + HF(g)
a. How many moles of each reactant are needed to produce 4.00 moles of HF?
b. How many grams of F2 are required to react with 1.50 moles of NH3?
c. How many grams of N₂F4 can be produced when 3.40 grams of NH3 reacts?
Answer:
2NH₃(g) + 5F₂(g) → N₂F₄(g) + 6HF(g)
(a) mol of NH₃ required = 1.333 mol; mol of F₂ required = 3.333 mol
(b) mass of F₂ required = 142.5 g
(c) N₂F₄ produced = 10.38 g
Explanation:
2NH₃(g) + 5F₂(g) → N₂F₄(g) + 6HF(g)
What is Stoichiometry?
In chemical equations, unless stated otherwise, the reactants and products will theoretically always remain in stoichiometric ratios.
The stoichiometry of a reaction is the relationship between the relative quantities of products and reactants, typically a ratio of whole integers.
Consider the following chemical reaction: aA + bB ⇒ cC + dD.
The stoichiometry of reactants to products in this reaction is the ratio of the coefficients of each species: a : b : c : d.
Converting between moles and mass:
To convert from mass to moles, divide the mass present by the molar mass, resulting in the number of moles.
Thence, the formula for moles: n = m/M, where n = number of moles, m = mass present, and M = molar mass. This formula can be easily rearranged to find mass present from molar mass and moles, or molar mass from mass and moles.
a. How many moles of each reactant are needed to produce 4.00 moles of HF?
In the given chemical equation, the stoichiometry of the reaction is
2 : 5 : 1 : 6. Therefore, for every 2 moles of NH₃, we require 5 moles of F₂, which will produce 1 mole of N₂F₄ and 6 moles of HF.
mol of NH₃ required = 1/3 × mol of HF = 1.333 mol
mol of F₂ required = 5/6 × mol of HF = 3.333 mol
b. How many grams of F₂ are required to react with 1.50 moles of NH₃?
Using stoichiometry again: mol of F₂ required = 5/2 × mol of NH₃
∴ F₂ required = 3.75 mol.
Then we can convert this to mass: m = nM = (3.75)(2×19.00) = 142.5 g
c. How many grams of N₂F₄ can be produced when 3.40 grams of NH₃ reacts?
Converting mass to moles: n = m/M = 3.40/(14.01+1.008×3) = 0.1996 mol
Using stoichiometry again: mol of N₂F₄ produced = 1/2 × mol of NH₃
∴ N₂F₄ produced = 0.0998 mol
converting moles to mass: m = nM = (0.0998)(14.01×2+19.00×4)
∴ N₂F₄ produced = 10.38 g
What is true of spontaneous reactions?
O They are indicated by a negative change in Gibbs free energy.
O They have a positive value of AS.
O They are instantaneous.
O They always release heat.
Help 20pts
An aqueous solution of 4mol/l nitric acid is electrolyzed in an electrolytic cell using graphite electrodes . write the chemical symbols for all the ions present in the electrolytic cell
The ions found in the acqueous solutions are:
H+, NO3-, O-
In the Periodic Table below, shade all the elements for which the neutral atom has an outer electron configuration of ms2nd2, where n and m are integers, and =m+n1.
The elements that have an outer electron configuration of ms2nd2 are located in the d-block of the periodic table and include some of the transition metals and lanthanides.
What is the periodic table?To determine which elements in the periodic table have this outer electron configuration, you can look at the position of the d-block elements in the table. The d-block elements are located in the middle of the table and include the transition metals. These elements have partially filled d orbitals, which can accommodate up to 10 electrons.
Elements in the d-block with an atomic number of 21 through 30 (scandium through zinc) have an outer electron configuration of d10s2 and do not fit the ms2nd2 configuration. However, elements in the d-block with an atomic number of 39 through 48 (yttrium through cadmium) have an outer electron configuration of d10s2p1 and can have the ms2nd2 configuration by removing the single electron in the p orbital. Elements in the d-block with an atomic number of 57 through 80 (lanthanum through mercury) also have the possibility of having an outer electron configuration of ms2nd2.
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